Download DB Systems - SELECT-INSERT-DELETE-UPDATE Practice - Individual

Faculty of Computer Science and Engineering
Ho Chi Minh City University of Technology
INDIVIDUAL PRACTICE ON
SELECT – CREATE VIEW INSERT – DELETE - UPDATE
Database Systems
(CO2013)
Computer Science Program
Dr. Võ Thị Ngọc Châu
([email protected])
Semester 1 – 2020-2021
Main references
Text:
 [1] R. Elmasri, S. R. Navathe, Fundamentals of Database
Systems- 6th Edition, Pearson- Addison Wesley, 2011.

R. Elmasri, S. R. Navathe, Fundamentals of Database Systems- 7th
Edition, Pearson, 2016.
References:
[1] S. Chittayasothorn, Relational Database Systems: Language,
Conceptual Modeling and Design for Engineers, Nutcha Printing Co.
Ltd, 2017.
[3] A. Silberschatz, H. F. Korth, S. Sudarshan, Database System
Concepts – 6th Edition, McGraw-Hill, 2006.
[4] H. G. Molina, J. D. Ullman, J. Widom, Database Systems: The
Complete Book - 2nd Edition, Prentice-Hall, 2009.
[5] R. Ramakrishnan, J. Gehrke, Database Management Systems - 2nd
Edition, McGraw-Hill.
[6] M. P. Papazoglou, S. Spaccapietra, Z. Tari, Advances in ObjectOriented Data Modeling, MIT Press, 2000.
[7]. G. Simsion, Data Modeling: Theory and Practice, Technics
Publications, LLC, 2007.
2
DML Statements

What you will do on the COMPANY database

Write SELECT statements for the queries and CREATE VIEW
statements for virtual tables

Write INSERT statements to populate data into the COMPANY
database

Write DELETE statements to remove data from the COMPANY
database


Write UPDATE statements to modify data in the COMPANY database
PREPARATION with the given files

Execute the DDL statements for the COMPANY database schema

Execute the INSERT statements to initialize the COMPANY database
3
The COMPANY Database
4
The COMPANY
database
Source: [1]
5
Practice 1










1. Return the department number and name of each department managed by
Smith, who does not live in Houston.
2. Return the department number and name of each department managed by
Smith and controlling projects in Houston.
3. Return the department number and name of each department in Houston but
not controlling any project in Houston.
4. Return the department number and name of each department that has the
highest number of employees.
5. Create a virtual table that keeps the details of each department in Houston:
department_number, department_name, number_of_projects, total_hours.
6. Create a virtual table that keeps the details of each department with:
dnumber, dname, no_of_employees, avg_salary.
7. Return the department number and name of each department that has the
less number of employees but higher average salary as compared to department
4.
8. Insert department 3, named Logistics, managed by employee 987987987,
from the same time when department Research has been managed.
9. Change the manager of department 3 to Smith, who lives in Houston.
10. Remove data of the department which has controlled only one project in
Houston.
6
Practice 1

1. Return the department number and name of each department managed by
Smith, who does not live in Houston.
SELECT dnumber, dname
FROM department
WHERE mgr_ssn IN (SELECT ssn
FROM employee
WHERE lname = 'Smith' AND NOT (address LIKE '%Houston%'));

2. Return the department number and name of each department managed by
Smith and controlling projects in Houston.
SELECT dnumber, dname
FROM department
WHERE mgr_ssn IN (SELECT ssn FROM employee WHERE lname = 'Smith')
AND dnumber IN (SELECT dnum FROM project WHERE plocation = 'Houston');
7
Practice 1

3. Return the department number and name of each department in Houston but
not controlling any project in Houston.
SELECT dnumber, dname
FROM department NATURAL JOIN dept_locations
WHERE dlocation = 'Houston'
AND dnumber NOT IN (SELECT dnum
FROM project
WHERE plocation = 'Houston');

4. Return the department number and name of each department that has the
highest number of employees.
SELECT dnumber, dname
FROM department JOIN employee ON dnumber = dno
GROUP BY dnumber
HAVING COUNT(*) = (SELECT MAX(cnt)
FROM (SELECT dno, COUNT(*) cnt
FROM employee
GROUP BY dno) dept_cnt);
8

5. Create a virtual table that keeps the details of each department in Houston:
department_number, department_name, number_of_projects, total_hours.
CREATE VIEW Houston_dept AS
SELECT dnumber department_number, dname department_name,
COUNT(DISTINCT pnumber) number_of_projects, SUM(Hours) total_hours
FROM department NATURAL JOIN dept_locations
JOIN project ON dnumber = dnum JOIN works_on ON pnumber = pno
WHERE dlocation = 'Houston'
GROUP BY dnumber, dname;

6. Create a virtual table that keeps the details of each department with:
dnumber, dname, no_of_employees, avg_salary.
CREATE VIEW dept_v AS
SELECT dnumber, dname, COUNT(*) no_of_employees, AVG(salary) avg_salary
FROM department JOIN employee ON dnumber = dno
GROUP BY dnumber, dname;

7. Return the department number and name of each department that has the
less number of employees but higher average salary as compared to department
4.
SELECT d1.dnumber, d1.dname
FROM dept_v d1 JOIN dept_v d4
ON d1.no_of_employees < d4.no_of_employees
AND d1.avg_salary > d4.avg_salary
WHERE d4.dnumber = 4;
9
Practice 1

8. Insert department 3, named Logistics, managed by employee 987987987, from
the same time when department Research has been managed.
INSERT INTO department (dnumber, dname, mgr_ssn, mgr_start_date)
VALUES (3, 'Logistics', '987987987', (SELECT mgr_start_date FROM department
WHERE dname = 'Research'));

9. Change the manager of department 3 to Smith, who lives in Houston.
UPDATE department
SET mgr_ssn = (SELECT ssn FROM employee WHERE lname = 'Smith'
AND address LIKE '%Houston%')
WHERE dnumber = 3;

10. Remove data of the department which has controlled only one project in
Houston.
DELETE FROM department
WHERE dnumber IN (SELECT dnum
FROM project
WHERE plocation = 'Houston'
GROUP BY dnum
HAVING COUNT(*) = 1);
10
Practice 1

8. Insert department 3, named Logistics, managed by employee 987987987, from
the same time when department Research has been managed.
INSERT INTO department (dnumber, dname, mgr_ssn, mgr_start_date)
VALUES (3, 'Logistics', '987987987', (SELECT mgr_start_date FROM department
WHERE dname = 'Research'));
Demonstration on MySQL
11
Practice 1

8. Insert department 3, named Logistics, managed by employee 987987987, from
the same time when department Research has been managed.
INSERT INTO department (dnumber, dname, mgr_ssn, mgr_start_date)
VALUES (3, 'Logistics', '987987987', (SELECT mgr_start_date FROM department
WHERE dname = 'Research'));
Demonstration on MySQL
12

8. Insert department 3, named Logistics, managed by employee 987987987,
from the same time when department Research has been managed.
Practice 1
INSERT INTO department (dnumber, dname, mgr_ssn, mgr_start_date)
VALUES (3, 'Logistics', '987987987', (SELECT mgr_start_date FROM department
WHERE dname = 'Research'));
INSERT INTO department (dnumber, dname, mgr_ssn, mgr_start_date)
VALUES (3, 'Logistics', '987987987', getMgrDate('Research'));
INSERT INTO department (dnumber, dname, mgr_ssn, mgr_start_date)
VALUES (13, 'IT', '987987987', (SELECT d.mgr_start_date
FROM department d JOIN department dd ON d.dnumber = dd.dnumber
WHERE d.dname = 'Research'));
Demonstration on MySQL
13
Practice 1

8’. Insert department 12, named Marketing, managed by employee 988988988,
from the same time when department Research has been managed.
INSERT INTO department (dnumber, dname, mgr_ssn, mgr_start_date)
VALUES (12, ‘Marketing', '988988988', (SELECT mgr_start_date FROM department
WHERE dname = 'Research'));
Msg 547, Level 16, State 0, Procedure insert_trigger, Line 19
The INSERT statement conflicted with the FOREIGN KEY constraint
"FK__DEPARTMEN__MGR_S__0DAF0CB0". The conflict occurred in database "company",
table "dbo.EMPLOYEE", column 'SSN'.
The statement has been terminated.
Demonstration on MS SQL Server
14
Practice 2










1. Retrieve the name, gender, and relationship of each dependent of the
employee in department 4.
2. Retrieve the name, gender, and date of birth of each dependent of the
manager of department Research.
3. Retrieve the details of the daughter of each employee who works on more
than 3 projects in Houston.
4. Retrieve the number of sons of each employee who lives in Houston and
works for departments also in Houston.
5. Create a virtual table that keeps the details of dependent of each employee
in department Research with: ssn, first_name, last_name, dependent_name,
gender, date_of_birth, relationship.
6. Create a virtual table that keeps the detailed number of dependents of each
employee with: ssn, first_name, last_name, department_name,
number_of_sons, number_of_daughters.
7. Retrieve ssn and name of each employee in department Research who has
more daughters than sons.
8. Insert Nancy as a daughter of the manager of department Research who was
born on July 12, 1990.
9. Change the name of the son of employee 543543543 to the first name of
his/her supervisor.
10. Remove the data of the dependents of each employee who works for the
department which has not yet been assigned any manager.
15
Practice 2

1. Retrieve the name, gender, and relationship of each dependent of the
employee in department 4.
SELECT dependent_name, sex, relationship
FROM dependent
WHERE essn IN (SELECT ssn FROM employee WHERE dno = 4);

2. Retrieve the name, gender, and date of birth of each dependent of the
manager of department Research.
SELECT dependent_name, sex, bdate
FROM dependent
WHERE essn IN (SELECT mgr_ssn FROM department WHERE dname = 'Research');

3. Retrieve the details of the daughter of each employee who works on more
than 3 projects in Houston.
SELECT *
FROM dependent
WHERE relationship = 'Daughter'
AND essn IN (SELECT essn FROM works_on JOIN project ON pno = pnumber
WHERE plocation = 'Houston'
GROUP BY essn
HAVING COUNT(*) > 3);
16
Practice 2

4. Retrieve the number of sons of each employee who lives in Houston and
works for departments also in Houston.
SELECT essn, COUNT(*) number_of_sons
FROM dependent
WHERE relationship = 'Son'
AND essn IN (SELECT ssn FROM employee
WHERE address LIKE '%Houston%' AND dno IN (SELECT dnumber
FROM dept_locations
WHERE dlocation = 'Houston'))
GROUP BY essn;

5. Create a virtual table that keeps the details of dependent of each employee
in department Research with: ssn, first_name, last_name, dependent_name,
gender, date_of_birth, relationship.
CREATE VIEW Research_dep AS
SELECT ssn, fname first_name, lname last_name, dependent_name, d.sex gender,
d.bdate date_of_birth, relationship
FROM dependent d JOIN employee ON essn = ssn
JOIN department ON dno = dnumber
WHERE dname = 'Research';
17
Practice 2

6. Create a virtual table that keeps the detailed number of dependents of each
employee with: ssn, first_name, last_name, department_name,
number_of_sons, number_of_daughters.
CREATE VIEW emp_dep AS
SELECT ssn, fname first_name, lname last_name, dname department_name,
number_of_sons, number_of_daughters
FROM department JOIN employee ON dnumber = dno
JOIN (SELECT essn, COUNT(*) number_of_sons FROM dependent
WHERE relationship = 'Son' GROUP BY essn) s ON ssn = s.essn
LEFT OUTER JOIN (SELECT essn, COUNT(*) number_of_daughters
FROM dependent
WHERE relationship = 'Daughter' GROUP BY essn) d ON ssn = d.essn;

7. Retrieve ssn and name of each employee in department Research who has
more daughters than sons.
SELECT ssn, first_name, last_name
FROM emp_dep
WHERE department_name = 'Research'
AND number_of_daughters > number_of_sons;
18
Practice 2

8. Insert Nancy as a daughter of the manager of department Research who was
born on July 12, 1990.
INSERT INTO dependent (essn, name, sex, bdate, relationship)
VALUES ((SELECT mgr_ssn FROM department WHERE dname = 'Research'),
'Nancy', 'F', '12-07-1990', 'Daughter');

9. Change the name of the son of employee 543543543 to the first name of
his/her supervisor.
UPDATE dependent
SET name = (SELECT fname FROM employee
WHERE ssn IN (SELECT super_ssn
FROM employee
WHERE ssn = '543543543'))
WHERE relationship = 'Son'
AND essn = '543543543';

10. Remove the data of the dependents of each employee who works for the
department which has not yet been assigned any manager.
DELETE FROM dependent
WHERE essn IN (SELECT ssn FROM employee
WHERE dno IN (SELECT dnumber FROM department
WHERE mgr_ssn IS NULL));
19
Practice 3










1. Retrieve ssn, first name, last name and department name of each
employee who works for the department that controls projects in Houston.
2. Retrieve ssn, first name, and last name of each employee who works on
projects each of which has more than 2 workers.
3. Retrieve ssn, first name, last name, salary, and department number of
each employee who has the highest salary in his/ her department.
4. Retrieve ssn, name, and department name of each employee who do not
work for any project in Houston.
5. Create a virtual table that keeps the details of each employee and his/ her
department with: ssn, first_name, last_name, department_number,
department_name
6. Create a virtual table that keeps the details of each employee and his/ her
projects with: ssn, first_name, last_name, project_number, project_name.
7. Retrieve ssn, department number, project number of each employee who
works on the same project Computerization as Wong but works in different
departments.
8. Insert data of employee 789789789 born on Feb 01, 1970, working for
the department where Wong’s supervisor works.
9. Update the salary of each new employee to the average salary of his/ her
department.
10. Remove data of each employee who has the total sum of hours working
on the projects less than 40.
20
Practice 3

1. Retrieve ssn, first name, last name and department name of each
employee who works for the department that controls projects in Houston.
SELECT ssn, fname, lname, dname
FROM employee JOIN department ON dno = dnumber
WHERE dnumber IN (SELECT dnum
FROM project
WHERE plocation = 'Houston');

2. Retrieve ssn, first name, and last name of each employee who works on
projects each of which has more than 2 workers.
SELECT ssn, fname, lname
FROM employee
WHERE ssn IN (SELECT essn
FROM works_on
WHERE pno IN (SELECT pno FROM works_on
GROUP BY pno
HAVING COUNT(*)>2));
21
Practice 3

3. Retrieve ssn, first name, last name, salary, and department number of
each employee who has the highest salary in his/ her department.
SELECT ssn, fname, lname, salary, dno
FROM employee e
WHERE salary = (SELECT MAX (salary) FROM employee WHERE dno = e.dno);

4. Retrieve ssn, name, and department name of each employee who do not
work for any project in Houston.
SELECT ssn, fname, lname, dname
FROM employee JOIN department on dno = dnumber
WHERE ssn NOT IN (SELECT essn
FROM works_on
WHERE pno IN (SELECT pnumber
FROM project
WHERE plocation = 'Houston'));
22

5. Create a virtual table that keeps the details of each employee and his/ her
department with: ssn, first_name, last_name, department_number,
department_name
CREATE VIEW emp_dept AS
SELECT ssn, fname first_name, lname last_name, dno department_number,
dname department_name
FROM employee JOIN department ON dno = dnumber;

6. Create a virtual table that keeps the details of each employee and his/ her
projects with: ssn, first_name, last_name, project_number, project_name.
CREATE VIEW emp_proj AS
SELECT ssn, fname first_name, lname last_name, pno project_number, pname
project_name
FROM employee JOIN works_on ON ssn = essn JOIN project ON pno = pnumber;

7. Retrieve ssn, department number, project number of each employee who
works on the same project Computerization as Wong but works in different
departments.
SELECT d.ssn, department_number, project_number
FROM emp_dept d JOIN emp_proj p ON d.ssn = p.ssn
WHERE project_name = 'Computerization'
AND department_number NOT IN (SELECT department_number
FROM emp_dept dw JOIN emp_proj pw ON dw.ssn = pw.ssn
WHERE project_name = 'Computerization'
AND dw.last_name = 'Wong');
23
Practice 3

8. Insert data of employee 789789789 born on Feb 01, 1970, working for
the department where Wong’s supervisor works.
INSERT INTO employee (ssn, bdate, dno)
VALUES ('789789789', '1970-02-01', (SELECT s.dno
FROM employee e JOIN employee s ON e.super_ssn = s.ssn
WHERE e.lname = 'Wong'));

9. Update the salary of each new employee to the average salary of his/ her
department.
UPDATE employee
SET salary = (SELECT AVG(salary) FROM employee e WHERE e.dno = dno)
WHERE salary IS NULL;

10. Remove data of each employee who has the total sum of hours working
on the projects less than 40.
DELETE FROM employee
WHERE ssn IN (SELECT essn
FROM works_on
GROUP BY essn
HAVING SUM(hours) < 40);
24
Database Systems (CO2013)
25